| Exam Board | CAIE |
|---|---|
| Module | P1 (Pure Mathematics 1) |
| Year | 2018 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Composite & Inverse Functions |
| Type | Find minimum domain for inverse |
| Difficulty | Moderate -0.8 This is a straightforward question on inverse functions requiring identification of the vertex for one-to-one domain (x ≥ 2), standard inverse function derivation by swapping and rearranging, and solving a composite function equation using substitution. All steps are routine textbook exercises with no novel problem-solving required. |
| Spec | 1.02u Functions: definition and vocabulary (domain, range, mapping)1.02v Inverse and composite functions: graphs and conditions for existence |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Smallest value of \(c\) is 2. Accept 2, \(c = 2\), \(c \geqslant 2\). Not in terms of \(x\) | B1 | Ignore superfluous working, e.g. \(\frac{d^2y}{dx^2} = 2\) |
| 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(y = (x-2)^2 + 2 \rightarrow x - 2 = (\pm)\sqrt{y-2} \rightarrow x = (\pm)\sqrt{y-2} + 2\) | M1 | Order of operations correct. Allow sign errors |
| \(f^{-1}(x) = \sqrt{x-2} + 2\) | A1 | Accept \(y = \sqrt{x-2} + 2\) |
| Domain of \(f^{-1}\) is \(x \geqslant 6\). Allow \(\geqslant 6\) | B1 | Not \(f^{-1}(x) \geqslant 6\). Not \(f(x) \geqslant 6\). Not \(y \geqslant 6\) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\left[(x-2)^2 + 2 - 2\right]^2 + 2 = 51\) SOI. Allow 1 term missing for M mark. Or \((x^2 - 4x + 6)^2 - 4(x^2 - 4x + 6) + 6 = 51\) | M1A1 | ALT. \(f(x) = f^{-1}(51)\) (M1) \(= \sqrt{51-2} + 2\) (A1) |
| \((x-2)^4 = 49\) or \((x^2 - 4x + 4)^2 = 49\). OR \(x^4 - 8x^3 + 24x^2 - 32x - 33 = 0\) often implied by next line | A1 | \((x-2)^2 + 2 = \sqrt{49} + 2\) OR \(f(x) = 9\) |
| \((x-2)^2 = (\pm)7\) OR \(x^2 - 4x - 3 = 0\). Ignore \(x^2 - 4x + 11 = 0\) | A1 | \((x-2)^2 = 7\) OR \(x = f^{-1}(9)\) |
| \(x = 2 + \sqrt{7}\) only CAO. \(x = 2 + \sqrt[4]{49}\) scores 3/5 | A1 | \(x = 2 + \sqrt{7}\) |
| 5 |
## Question 10(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Smallest value of $c$ is 2. Accept 2, $c = 2$, $c \geqslant 2$. Not in terms of $x$ | B1 | Ignore superfluous working, e.g. $\frac{d^2y}{dx^2} = 2$ |
| | **1** | |
## Question 10(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $y = (x-2)^2 + 2 \rightarrow x - 2 = (\pm)\sqrt{y-2} \rightarrow x = (\pm)\sqrt{y-2} + 2$ | M1 | Order of operations correct. Allow sign errors |
| $f^{-1}(x) = \sqrt{x-2} + 2$ | A1 | Accept $y = \sqrt{x-2} + 2$ |
| Domain of $f^{-1}$ is $x \geqslant 6$. Allow $\geqslant 6$ | B1 | Not $f^{-1}(x) \geqslant 6$. Not $f(x) \geqslant 6$. Not $y \geqslant 6$ |
| | **3** | |
## Question 10(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\left[(x-2)^2 + 2 - 2\right]^2 + 2 = 51$ SOI. Allow 1 term missing for M mark. Or $(x^2 - 4x + 6)^2 - 4(x^2 - 4x + 6) + 6 = 51$ | M1A1 | ALT. $f(x) = f^{-1}(51)$ (M1) $= \sqrt{51-2} + 2$ (A1) |
| $(x-2)^4 = 49$ or $(x^2 - 4x + 4)^2 = 49$. OR $x^4 - 8x^3 + 24x^2 - 32x - 33 = 0$ often implied by next line | A1 | $(x-2)^2 + 2 = \sqrt{49} + 2$ OR $f(x) = 9$ |
| $(x-2)^2 = (\pm)7$ OR $x^2 - 4x - 3 = 0$. Ignore $x^2 - 4x + 11 = 0$ | A1 | $(x-2)^2 = 7$ OR $x = f^{-1}(9)$ |
| $x = 2 + \sqrt{7}$ only CAO. $x = 2 + \sqrt[4]{49}$ scores 3/5 | A1 | $x = 2 + \sqrt{7}$ |
| | **5** | |10 The one-one function f is defined by $\mathrm { f } ( x ) = ( x - 2 ) ^ { 2 } + 2$ for $x \geqslant c$, where $c$ is a constant.\\
(i) State the smallest possible value of $c$.\\
In parts (ii) and (iii) the value of $c$ is 4 .\\
(ii) Find an expression for $\mathrm { f } ^ { - 1 } ( x )$ and state the domain of $\mathrm { f } ^ { - 1 }$.\\
(iii) Solve the equation $\mathrm { ff } ( x ) = 51$, giving your answer in the form $a + \sqrt { } b$.\\
\hfill \mbox{\textit{CAIE P1 2018 Q10 [9]}}